Silly dicey problem

[Batalov]

Originally Posted by wblipp

In my experience, nobody still arguing is likely to be convinced. They will fall into three categories:
1. Those so cocksure of their answer that they don't think about, and probably don't even read, your explanation.
2. Those so mathematically naive they don't really understand the question.
3. Trolls.

If it were me, I would post the clearest explanation I could compose and then withdraw from the discussion.

This is a great recipe for any argument in modern life! Very wise!

It is applicable to discussing anything:
- global warming
- capitalism = freedom
- 9/11
- all is quiet on the Eastern front
- downing of MH17
- are you a patriot and "if you are then why are you saying what you are saying, you moron?!"
- math paradoxes
- arsenic life
- high school kid geniuses robbed of their inventions by corporations
- you name it
- ...and definitely religion

[R. Gerbicz]

Originally Posted by sdbardwick

What are the odds that both of the cubes show a 2?
[/SPOILER]

Depends on the partner's strategy. This can be (almost) everything. For example he says two only if you have thrown 2-2, otherwise selects the cube that is different from two (if both of them are different from two, then randomly). In this case if he says two, then the other cube is also two. So the probability is exactly 1.

Hence I would say that the probability doesn't exist if we don't know the partner's strategy.

[sdbardwick]

Originally Posted by R. Gerbicz

Depends on the partner's strategy. This can be (almost) everything. For example he says two only if you have thrown 2-2, otherwise selects the cube that is different from two (if both of them are different from two, then randomly). In this case if he says two, then the other cube is also two. So the probability is exactly 1.

Hence I would say that the probability doesn't exist if we don't know the partner's strategy.

No partner strategy. He can only truthfully report "at least one die shows a 2", otherwise he is mute.

Thread I lifted this from made this clear; sorry I didn't.

[science_man_88]

Originally Posted by henryzz

A slight modification to the question:

You throw three fair six-sided dice simultaneously so that each lands by itself in one of three separate boxes. You cannot see into the boxes, but your partner can. Your partner then truthfully tells you that at least two of the dice are showing a 2.

What are the odds that all three of the cubes show a 2?

Please explain the process used to arrive at your answer.

Can anyone work out a formula for the generalization to n dice with n-1 known to be showing a 2 or better still with x known to be showing a 2?

well in theory on a dice with x faces we get that there's (n n-1) ( n choose n-1) combinations of n-1 that are the same in n total dice the odds of landing on any one number in theory is 1/x so we get one in x*(n n-1) but we have to subtract the duplicates with all them the same there are (n n-1)-1 duplicates so we get one in (x-1)*(n n-1)+1 I think. edit: though I see why some may say 1/6 edit2: ordering of the duplicates are different so going to permutations changes the answer

[LaurV]

This is the boy/girl paradox, version 2. The tricky part is "at least one". The correct solution is given in first two posts of the topic (yes, I have read the spoilers before trying to answer, but I am sure I could find the answer).

edit: waaa... a lot of talking! it put me on the second page already!
edit2: nice explanation and very good post wblipp!

[retina]

If you offer 9:1 odds (i.e. sweeten the pot to appear to be a 10/6 return EV for the mathematically challenged) you might get more takers. And the 91% return ratio is worse (or better depending upon your POV) than most casino games.

[sdbardwick]

Darn, the second-most strident supporter of 1/6 just changed his mind. And he managed to (inadvertently) sort of poach my idea on the betting game. My revised plan was to roll the dice (like in a backgammon dice cup, but leave the cup inverted over the roll), have a 3rd party (the spotter) look at the dice, and announce one of the numbers showing. The bettor could then bet on if it was a pair, and either lose 1 unit if it was not a pair, or get paid 8 units if it was a pair. No cheating by the spotter, as both dice would be exposed before the next roll. Speeds up the game as you don't need to wait for a specific number to show...the former believer envisioned the same thing, but substituted an optical reader for the human spotter.

Although there could be collusion between the spotter and bettor...so it would have to be the backer (me) announcing the number, I guess.

[LaurV]

[offtopic]

Originally Posted by wblipp

None of these ideas has ever worked for me on the Monty Hall problem...

What was (occasionally) working for me on MH problem was "imagine you have 100 (then 10) doors and one have a car, what is your chance to pick the car on a single try?" For this, the most interlocutors identify the correct, 1/100 (then 1/10) answer. "So, the probability that the car is behind one of the other 99 rooms is 99/100", to which they agree. Then "now, I, the host, tell you 'do you want to stay to your initial pick (1 door), or do you want to pick the other 99 doors, all together? If one of them contains the car, it is yours'". The fact that the host opens 98 doors with goats, or opens 5, or none, makes no difference. This convinces the most of the people. Then I make it smaller, 10 rooms as said in parenthesis, 5 rooms, 3 rooms. Most of them are convinced they must switch, even before I get to 3 rooms.
[/offtopic]

[retina]

Originally Posted by LaurV

... MH problem ...

Who wants a stupid car anyway?

[davar55]

Originally Posted by wblipp

In my experience, nobody still arguing is likely to be convinced. They will fall into three categories:
1. Those so cocksure of their answer that they don't think about, and probably don't even read, your explanation.
2. Those so mathematically naive they don't really understand the question.
3. Trolls.

If it were me, I would post the clearest explanation I could compose and then withdraw from the discussion. My explanation would be the following diagram. It says the same thing others on this thread have said in words, but a picture has a small chance of reaching some additional people by accessing a different way of thinking.

   1 2 3 4 5 6
  +-----------
1 |  X
2 |X X X X X X
3 |  X
4 |  X
5 |  X
6 |  X

I know some wonderful and clever ideas about how to engage the first group. None of these ideas has ever worked for me on the Monty Hall problem, so they must be wonderful and clever in my mind, not in the real world.

D*mn, I wish I'd thought of actually typing up that solution/explanation. It makes it so clear. Nice.

[TheMawn]

Here's an interesting monkey wrench...

If you're not limiting yourself to 2's in order to make the game go faster then there's some human non-randomness that gets thrown into the mix, too. For example, if the caller says "One of the dice is a 3" the player wins if they're both 3. It would be like this as opposed to waiting for specifically a 2.

How does the caller decide which die to consider? For example, if you get a 2 and a 5, the caller could say "One of the dice is a 2" or "One of the dice is a 5," right? If he always goes with the left-most one, for example, then the odds become 1 in 6.